$\dfrac{2}{6} + \dfrac{1}{3} = {?}$
Solution: ${\dfrac{2}{6}}$ ${\dfrac{1}{3}}$ $+$ ${\dfrac{2 \times 1}{6 \times 1}}$ ${\dfrac{1 \times 2}{3 \times 2}}$ $+$ ${\dfrac{2}{6}}$ ${\dfrac{2}{6}}$ $+$ $ = \dfrac{{2} + {2}}{6} $ $ = \dfrac{4}{6}$